The set of matrices that are their own inverse in R2

3. The attempt at a solution
I know that the diagonal is comprised of 1s and or -1s and the other entries are zero but I can't seem to show it algebraically.

I went the weak way and did components....
I said A had rows of {a,b} and {c,d} and that A2=I
so
a2+bc=1
b(a+d)=0
c(a+d)=0
d2+bc=1

Now I know that the solution will have diagonals comprised of (+ or -)1 and the other two entries zero but I don't know how to do the algebra to get there =| I'll write down what I tried last night, any direction would be greatly appreciated!

So I said that
b(a+d)=0
c(a+d)=0
so either (a+d)=0 or b=0 and c=0...
so if we assume a+d=!0 then
b=0 and c=0
thus d2=1
and a2=1
so we'd have a=+-1,b=0,c=0,d=+-1

but if a+d=0
then a=-d
d2-a2=0
d=+-a but apparently the positive case is ignored since we assumed a+d=0 s0 d=-a again and I have nothing conclusive about b or c?

Again, thank you in advanced for any and all advice....it's sad that my basic algebra skills are so weak....if someone could recommend a good algebra practice book that would also be appreciated. I think I'm worse at Algebra than everything else haha

Also if this thread needs to be moved just let me know, at my university Linear Algebra is beyond Calculus but this particular question is fairly simple.